我已经将apache wink rest servlet用于REST服务,并且我试图将其与Spring MVC调度程序servlet集成。但是调度程序Servlet的网址无法访问。 这是我的web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="3.0"
xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd">
<display-name>DomesticCollect</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<!-- it appears that below "classpath:META-INF/server/wink-core-context.xml"
is needed for WebSphere 7 ; this is apparently not needed for 8 -->
<!-- classpath:META-INF/server/wink-core-context.xml -->
<context-param>
<param-name>contextConfigLocation</param-name>
<!-- <param-value> classpath:WEB-INF/ApplicationContext-context.xml</param-value> -->
<!-- com/qvc/supplychain/app/service/web/config/ -->
<param-value>
classpath:ApplicationContext-context.xml
</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>DomesticCollectService</servlet-name>
<servlet-class>org.apache.wink.server.internal.servlet.RestServlet</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.qvc.supplychain.app.service.web.controller.IntegrationService</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>DispatcherServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<servlet>
<servlet-name>DispatcherServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>DispatcherServlet</servlet-name>
<url-pattern>/admin/*</url-pattern>
</servlet-mapping>
<!-- <resource-ref>
<description>Connection to Main DB</description>
<res-ref-name>jdbc/SCMPOFRT DB2 DataSource</res-ref-name>
<res-type>javax.sql.DataSource</res-type>
<res-auth>Container</res-auth>
<res-sharing-scope>Shareable</res-sharing-scope>
</resource-ref>
<resource-ref>
<description>Connection to secondary DB</description>
<res-ref-name>jdbc/tmsVendorAsnDS</res-ref-name>
<res-type>javax.sql.DataSource</res-type>
<res-auth>Container</res-auth>
<res-sharing-scope>Shareable</res-sharing-scope>
</resource-ref> -->
</web-app>
这是我的DispatcherServlet-servlet.xml:
<bean id="urlMapping" class="org.springframework.web.servlet.handler.SimpleUrlHandlerMapping">
<property name="alwaysUseFullPath">
<value>true</value>
</property>
<property name="mappings">
<props>
<prop key="/admin/**">ApplicationPropertyController</prop>
</props>
</property>
</bean>
<bean id="ApplicationPropertyController" class="com.qvc.supplychain.app.domesticcollect.web.ApplicationPropertyController">
<property name="applicationPropertyManager" ref="ApplicationPropertyManager" />
</bean>
<mvc:annotation-driven/>
<!-- <context:component-scan base-package="com.qvc.supplychain.app.domesticcollect.web"/> -->
<bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="viewClass"><value>org.springframework.web.servlet.view.JstlView</value></property>
<property name="prefix" value="/WEB-INF/jsps/"/>
<property name="suffix" value=".jsp"/>
</bean>
如何提供url模式,以便可以访问其余服务和调度程序servlet?